A shell in flight explodes into n equal fragments, k of the fragments reach the ground earlier than the other fragments. The acceleration of their centre of mass subsequently will be

# A shell in flight explodes into n equal fragments, k of the fragments reach the ground earlier than the other fragments. The acceleration of their centre of mass subsequently will be

1. A

g

2. B

(n-k)g

3. C

$\frac{\left(\mathrm{n}-\mathrm{k}\right)\mathrm{g}}{\mathrm{k}}$

4. D

$\frac{\left(\mathrm{n}-\mathrm{k}\right)}{\mathrm{n}}\mathrm{g}$

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### Solution:

If 'k' fragments reaches ground

So, their acceleration = 0

${\mathrm{a}}_{\mathrm{CM}}=\frac{{\mathrm{m}}_{1}{\mathrm{a}}_{1}+.....+{\mathrm{m}}_{2}{\mathrm{a}}_{2}}{{\mathrm{m}}_{1}+....+\mathrm{mn}}$

$=\frac{\left(\mathrm{n}-\mathrm{k}\right)\mathrm{mg}}{\mathrm{nm}}=\frac{\left(\mathrm{n}-\mathrm{k}\right)\mathrm{g}}{\mathrm{n}}$

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